Lecture 2 Analog circuits. IR detection
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1 Seeing the light.. Lecture Analog circuits I t IR light V 9V V Q OP805 RL IR detection Noise sources: Electrical (60Hz, 0Hz, 80Hz.) Other electrical IR from lights IR from cameras (autofocus) Visible light ~ mv What we want: 0 5 V DC signal representing the IR amplitude. t
2 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect Analog circuits discrete devices: BJT Application: light detection Phototransistor: Acts like BJT except charge carriers generated by incident light add to the base current. In other words, I c ;Incident light V 9V V Q OP805 RL I c
3 Analog circuits DC block Capacitors: Open circuits for DC Short circuit at high frequencies Analog circuits: Op-amps V Ideal op-amp: V- - -Vcc Vcc = K (V -V - ) K ~ 0 6 (at DC!) With feedback to limit gain: V V- - V - = V out V out = K (V -V out ) When negative feedback is applied to an op-amp, V = V - if the amplifier is within its operating range. V out (K) = KV V out =V (K/(K)) V out = V 3
4 Analog circuits: Op-amps Eg: Inverting amplifier. Z V- - V - = 0 I = V in/ Z I Eg : Z = 00kΩ Z = 0kΩ Z V out = - 0 V in V out = 0 Z I V out = - (Z /Z ) V in Eg : Z = 00kΩ Z = Ω V out = - 00,000 V in!! Not likely. Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Z V- - Output current limitations I Z Since V - is a virtual ground, input impedance seen by V in is Z 4
5 Things to consider: Input impedance Gain Bandwidth product Bias Currents Analog circuits: Real Op-amps Voltage limitations Output current limitations V- - Since Op-amp inputs source or sink very little current (depends on type), input impedance in this case is very high. This is a commonly used buffer to separate your low impedance circuit from a sensitive source that you need to measure without drawing current. 5
6 Things to consider: Input impedance Gain Bandwidth product Bias Currents Analog circuits: Real Op-amps Voltage limitations Output current limitations Slew-rate is a similar limit: it is a limit on the rate of change of output voltage Z I Open loop gain (K) 0 V- 00 khz - Z Frequency Gain-Bandwidth limit (Hz) = Gain * Max. Frequency = CONSTANT 6
7 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Z V- - Output current limitations I Op-amp terminals can act as small current sources. These Bias Currents can become large error or offset voltages if the resistors in the circuit are large. Z Eg: 0 na bias current * 0 MΩ = 00 mv! 7
8 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current limitations Op-amp input voltages (V, V - ) must be at least a few volts away from the power rails (Vcc, -Vcc). Applying input voltages equal or near the power rails will cause the Op-amp to behave unexpectedly. V V- Rail-to-rail Op-amps are an expensive solution to this limitation. - Vcc -Vcc 8
9 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current / voltage limitations Op-amp output terminals can only provide a few ma of current. Motors, lamps and similar high current devices cannot typically be driven by a normal OPamp. High power Op-amps exist that can provide much higher current levels. Output voltage range is also limited within a few volts of the power rails. V V- - Vcc -Vcc 9
10 Analog circuits: Real Op-amps Application: amplifier stages U TL08 R3 30k R5 30k R6 k 0k R k R TL08 U k R4 TL08 U3 Total gain: 30*30* = 9900 Input impedance = high Max Bandwidth = 00 khz 0
11 Analog circuits: Filters Before you understand filters you must understand the difference between the TIME DOMAIN and FREQUENCY DOMAIN V o l t s 5.00 Scroll Right ---> for more hardware settings.00 A C Q U I S I T I O N S E T T I N G S C h a n n e l D e v i c e V r m s.49 : Peak Frequency : Peak Power Vrms^ Hz s e c 0 S a m p l i n g R a t e ( H z ) 3.0 F r a m e S i z e W i n d o w N o n e ( U n i f o r m ).0 A v e r a g e s H z D I S P L A Y S E T T I N G S A c q u i r e O N A c q u i r e O N C E S T O P V r m s L i n e a r Analog circuits: Filters Transfer Function = / = H(ω) So: V out (ω) = H(ω)*V in (ω) This is all in terms of ω since, in general, impedances are functions of ω. Z cap = /j ωc Z ind = j ωl Z res = R Similar to voltage divider: except ω dependent. Frequency Generator V in Z V out Z Spectrum Analyzer ground
12 Analog circuits: Filters V out (ω) = [V in (ω)/(ζ Ζ )] Ζ So: H(ω) = Z /(Z Z ) For resistors, this is just the well known voltage divider: R /(R R ) Frequency Generator V in Z V out Z Spectrum Analyzer ground Analog circuits: Filters V out (ω) = [V in (ω)/(ζ Ζ )] Ζ = So: H(ω) = Z /(Z Z ) / jωc R / jωc Now plug in a resistor and a capacitor: Z cap = /j ωc Z ind = j ωl Z = jωrc Frequency Generator R 3300 k C nf nf Spectrum Analyzer Z
13 Analog circuits: Filters For low frequencies (small ω), H = = For high frequencies (large ω), Η = 0 / jωc R / jωc At ω = j/rc, (the pole of this function) H is infinite. At the real value ω = /RC, H begins to decrease in amplitude. Z = jωrc Frequency Generator R 3300 k C nf nf Spectrum Analyzer f 0 = /(πrc) = 30 Hz Z Analog circuits: Filters Bode plots: a graphical representation of frequency response on logarithmic axes. Vertical axis: dbv = 0log 0 ( V ) Horizontal axis: log 0 (f) (0 is used instead of 0 so the result will represent power ~ V ) -3 db = ½ as much power as 0 db Log of frequency is used to ensure linear plots from /f or /f n functions Pole: /(jω/ω 0 ) -0 db/decade in amplitude after ω 0, -90 phase Zero: (jω/ω 0 ) 0 db/decade in amplitude after ω 0, 90 phase 3
14 Bode Plot: M a g n i t u d e ( d B ) Analog circuits: Filters = jωrc w 0 ( H z ) M a x f r e q ( H z ) P h a s e ( d e g r e e s ) dB, 30 Hz -0db/decade / pole - 45 deg, 30 Hz -90 deg / pole Analog circuits: Filters = jωrc Ch0 Ch R 3300 k C nf nf f 0 = /(πrc) = 30 Hz Gnd 4
15 Analog circuits: Active Filters Active Band Pass: R R C C U TL08 Z Z = ( jωc) R Z H(ω) = - Z /Z Z = R jωc Active Band Pass: Analog circuits: Active Filters R C Z R C U TL08 Z Zero at ω=0 = ( jωrc jωr C )( jωr C ) Pole at ω=/(r C ) Pole at ω=/(r C ) 5
16 Analog circuits: Active Filters 0log( H ) Idealized Bode Plot: 0log(R /R ) db () () () () () (3) 0db/decade 0 db () /(R C ) /(R C ) Zero at ω=0 = ( jωrc jωr C )( jωr C ) ω () (3) Pole at ω=/(r C ) Pole at ω=/(r C ) phase(h) Analog circuits: Active Filters Idealized Bode Plot: () () () -70 /(R C ) /(R C ) Zero at ω=0 () jωrc = ( jωr C )( jωr C ) () () (3) ω () (3) Pole at ω=/(r C ) Pole at ω=/(r C ) 6
17 Analog circuits filtering and detection Band Pass IR DC Amplify Filter detect block R R R C C C Low pass R C U TL08 R C Use multiple stages to get steeper filter roll-offs U TL08 R C U TL08 H tot (ω) = H (ω) * H (ω)*h 3 (ω) Remember 0dB/dec for each POLE 7
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========================================================================================== UNIVERSITY OF SOUTHERN MAINE Dept. of Electrical Engineering TEST #3 Prof. M.G.Guvench ELE343/02 ==========================================================================================
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